cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A227528 a(n) = numerator of r(n) = 2*(3*n+2)!/((2*n)!*2^n), n>=0.

This page as a plain text file.
%I A227528 #31 Jul 05 2017 21:11:05
%S A227528 4,60,840,13860,270270,6126120,158722200,4633467300,150587687250,
%T A227528 5394582443250,211240491462000,8977720887135000,411608985890602500,
%U A227528 20251162105817643000,1064311075116860571000,59509669251792738478500,3527387653231263127556250,220942735734212754080568750
%N A227528 a(n) = numerator of r(n) = 2*(3*n+2)!/((2*n)!*2^n), n>=0.
%C A227528 The first values with denominators > 1 occur at n = {84, 148, 164, 168, 169, 276, 292, 296, 297, ...}. - _G. C. Greubel_, Jul 04 2017
%H A227528 G. C. Greubel, <a href="/A227528/b227528.txt">Table of n, a(n) for n = 0..370</a>
%F A227528 In Maple notation,
%F A227528 E.g.f. of r: ((135/8)*z+4)*cos((2/3)*arcsin((3/4)*sqrt(6)*sqrt(z)))/(1-(27/8)*z)^2+(3/8)*sqrt(z)*((135/4)*z+17)*sin((2/3)*arcsin((3/4)*sqrt(6)*sqrt(z)))*sqrt(6)/(1-(27/8)*z)^(5/2).
%F A227528 E.g.f. of r: 4*hypergeom([4/3,5/3],[1/2],27*z/8).
%F A227528 Integral representation as n-th moment of a signed function w(x) of bounded variation on (0,infinity),
%F A227528   w(x)=(8/3)*sqrt(3)*((1/36) *(128*x^2/81-40*x/3+20) *exp(-4*x/27) *BesselK(1/3,(4/27)*x)/Pi +(2/81)*x *(-5+16*x/9) *exp(-4*x/27) *BesselK(4/3,4*x/27)/Pi);
%F A227528   w(x)=-(14/243)*16^(2/3)*x^(2/3)*3 *hypergeom([13/6], [4/3], -(8/27)*x)/GAMMA(2/3)-(10/243)*sqrt(3)*16^(1/3)*x^(1/3)*9*GAMMA(2/3)*hypergeom([11/6], [2/3], -(8/27)*x)/Pi.
%F A227528   For x>3.32, w(x)>0.
%F A227528   w(0)=w(3.32)=limit(w(x),x=infinity)=0.
%F A227528   For x<3.32, w(x)<0.
%F A227528 r(n) = int(x^n*w(x), x=0..infinity), n>=0.
%F A227528 Asymptotics: r(n)->(1/1152)*sqrt(6)*(10368*n^2+10224*n+2161)*(27/8)^n*exp(-n)*(n^n), for n->infinity.
%F A227528 The rational values are given by 4*(-2*n+1)*r(n) + 3*(3*n+2)*(3*n+1) * r(n-1)=0. - _R. J. Mathar_, Jul 20 2013
%p A227528 seq(numer(4*(3*n+2)!/((2*n)!*2^(n+1))), n=0..14);
%t A227528 Table[Numerator[2(3n + 2)!/((2n)! 2^n)], {n, 0, 84}]  (* _G. C. Greubel_, Jul 04 2017 *)
%o A227528 (PARI) for(n=0,50, print1(numerator(2*(3*n + 2)!/((2*n)!*2^n)), ", ")) \\ _G. C. Greubel_, Jul 04 2017
%Y A227528 Cf. A001879.
%K A227528 nonn,frac
%O A227528 0,1
%A A227528 _Karol A. Penson_, Jul 14 2013
%E A227528 Erroneous definition corrected by _G. C. Greubel_, Jul 04 2017